If $f:X \to Y$ induces isomorphism in cohomology, then it induces isomorphism in homology

algebraic-topologyhomological-algebrahomology-cohomology

Let $f:X \to Y$ and $G$ a group (I'm interested specifically in the case $G=\mathbb{Q}$) such that $f^\ast:H^q(X;G) \to H^q(Y;G)$ is an isomorphism for all $q$.
It it true that $f_*:H_q(X;G) \to H_q(Y;G)$ for all $q$?

I know that some sort of converse is true: if $f:X \to Y$ induces $f_*:H_q(X) \to H_q(Y)$ isomorphisms for all $q$, than by the Universal Coefficent Theorem we have that $f^\ast:H^q(X;G) \to H^q(Y;G)$ are isomorphism for all $q$ and for all $G$ groups.

In this other question I got a comment from Qiaochu Yuan that suggested that a map induces an isomorphism on rational cohomology iff it induces an isomorphism on rational homology, suggesting that it follows from the Universal Coefficient Theorem, but I don't know how.

Any help is appreciated!

Best Answer

Alessandro has explained the UCT part of the argument but I want to explain the $\mathbb{Q}$-linear dual part a bit more carefully because there is a mild subtlety in it in the infinite-dimensional case. The claim at stake is the following:

Conservativity of the dual: Let $f : V \to W$ be a linear map between vector spaces (not necessarily finite-dimensional) over a field $K$. Then $f$ is an isomorphism if and only if the dual map $f^{\ast} : W^{\ast} \to V^{\ast}$ is an isomorphism.

Note that this is not true for modules over a non-field (e.g. over $\mathbb{Z}$), and that to apply it to the original claim about inducing isomorphisms in homology vs. cohomology we need to know that the identification $H^i(X, \mathbb{Q}) \cong H_i(X, \mathbb{Q})^{\ast}$ is natural (which is true since the UCT itself is natural).

Proof. In the finite-dimensional case, conservativity of the dual follows from the fact that the double dual of $V$ is naturally isomorphic to it, so if $f^{\ast}$ is an isomorphism then so is $(f^{\ast})^{\ast} \cong f$. In the infinite-dimensional case this is very far from true. But without taking the double dual, we can instead just argue that if $f^{\ast}$ is surjective then $f$ is injective and vice versa:

  • Taking contrapositives, if $f$ is not injective then $f(v) = 0$ for some nonzero $v \in V$. Then if $v^{\ast} \in V^{\ast}$ is any linear functional such that $v^{\ast}(v) = 1$ (which exists assuming the axiom of choice!), it follows that $\text{im}(f^{\ast})$ does not contain $v^{\ast}$, so $f^{\ast}$ is not surjective.
  • Again taking contrapositives, if $f$ is not surjective then $W/\text{im}(f)$ is nonzero. Then if $w^{\ast} \in W^{\ast}$ is any linear functional which vanishes on $\text{im}(f)$ (equivalently, any linear functional on $W/\text{im}(f)$) but which does not vanish identically (which, again, exists assuming the axiom of choice!), it follows that $f^{\ast}(w^{\ast}) = 0$, so $f^{\ast}$ is not injective.

We conclude that if $f^{\ast}$ is an isomorphism then $f$ is both injective and surjective, hence an isomorphism. $\Box$

Now for the

Mild subtlety: In both arguments we needed to know that vector spaces have "enough linear functionals": the key fact we need is that linear functionals separate points in the sense that if $v \in V$ is nonzero then there exists a linear functional $v^{\ast} \in V^{\ast}$ which does not vanish on it (which one might call the "algebraic Hahn-Banach theorem"). This is obvious in the finite-dimensional case by extending $v$ to a basis, and it continues to be true in the infinite-dimensional case for the same reason, but now we need the axiom of choice to extend $v$ to a basis.

In the absence of the axiom of choice this can fail badly, and it can happen that the $\mathbb{Q}$-linear dual of a $\mathbb{Q}$-vector space is zero! ($\mathbb{R}$ is an example.) This would make the desired statement no longer true, if we take very strange examples for $X$ such as Eilenberg-MacLane spaces $K(\mathbb{R}, 1)$.

Fortunately I don't think this really affects the intended application. First of all I think all of the rational homology in the intended application is finite-dimensional. But even if it weren't, if the spaces involved have reasonable CW decompositions then one ought to be able to write down more-or-less explicit bases of rational homology using cells even in the infinite-dimensional case and then I think there's no problem.

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